Answer:
The average energy delivered to a surface is 19.116 W/m².
Explanation:
Given;
maximum electric field, E₀ = 120 v/m
The average energy delivered by the wave to a surface is given by
[tex]I_{avg} = \frac{c\epsilon_ o E_o^2}{2}[/tex]
where;
c is the speed of light, = 3 x 10⁸ m/s
ε₀ is the permittivity of free space = 8.85 x 10⁻¹² c²/Nm²
[tex]I_{avg} = \frac{c\epsilon_ o E_o^2}{2} \\\\I_{avg} = \frac{(3*10^8)(8.85*10^{-12})( 120)^2}{2}\\\\ I_{avg} =19.116 \ W/m^2[/tex]
Therefore, the average energy delivered to a surface is 19.116 W/m².
you have one 100 ohm resistor what size resistor must you hook up in parallel in order to have a resistance of 99.0
Answer:
The resistor to be hooked up must be 9900 [tex]\Omega[/tex]
Explanation:
Recall how resistors in parallel combine to render an equivalent resistor:
[tex]\frac{1}{R_e} =\frac{1}{R_1} +\frac{1}{R_2}[/tex]
So we need to find which resistor (R2) we need to combine in parallel with resistor R1 (100 [tex]\Omega[/tex]) in order to obtain an equivalent resistor Re (99 [tex]\Omega[/tex]) :
[tex]\frac{1}{99} =\frac{1}{100} +\frac{1}{R_2}\\\frac{9900\,R_2}{99} =\frac{9900\,R_2}{100} +\frac{9900\,R_2}{R_2}\\100\,R_2= 99\,R_2+9900\\R_2=9900\,\,\Omega[/tex]
Amy runs 2 blocks south, then turns around and runs 3 blocks north. what is the distance and displacement
Answer: Distance= 5 blocks and displacement= -1 block
Explanation: distance is the total length of a journey while displacement is the shortest straight length of a journey!
If i push a ball along the floor, which forces are acting upon the ball? Name atleast two forces and explain them !
Answer:
Hey there!
You are applying a force, pushing the ball, which makes the ball move forwards.
Friction is working in the opposite direction, slowing the ball down as it rolls on the floor.
Let me know if this helps :)
What is the magnitude of the total acceleration of point A after 2 seconds? The bar starts from rest and has a constant angular acceleration of 1.0 rad/s2 clockwise.
Answer:
a_total = 2 √ (α² + w⁴) , a_total = 2,236 m
Explanation:
The total acceleration of a body, if we use the Pythagorean theorem is
a_total² = a_T²2 + [tex]a_{c}[/tex]²
where
the centripetal acceleration is
a_{c} = v² / r = w r²
tangential acceleration
a_T = dv / dt
angular and linear acceleration are related
a_T = α r
we substitute in the first equation
a_total = √ [(α r)² + (w r² )²]
a_total = 2 √ (α² + w⁴)
Let's find the angular velocity for t = 2 s if we start from rest wo = 0
w = w₀ + α t
w = 0 + 1.0 2
w = 2.0rad / s
we substitute
a_total = r √(1² + 2²) = r √5
a_total = r 2,236
In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m
a_total = 2,236 m
A car traveling initially at 8.0 m/s accelerates to a velocity of 14 m/s in 2.0s. What is the average acceleration of the car?
Answer:
3.0 m/s²
Explanation:
Given:
v₀ = 8.0 m/s
v = 14 m/s
t = 2.0 s
Find: a
v = at + v₀
14 m/s = a (2.0 s) + 8.0 m/s
a = 3.0 m/s²
A wheel that is rotating at 33.3 rad/s is given an angular acceleration of 2.15 rad/s 2. Through what angle has the wheel turned when its angular speed reaches 72.0 rad/s
Answer:
The angle through which the wheel turned is 947.7 rad.
Explanation:
initial angular velocity, [tex]\omega _i[/tex] = 33.3 rad/s
angular acceleration, α = 2.15 rad/s²
final angular velocity, [tex]\omega_f[/tex] = 72 rad/s
angle the wheel turned, θ = ?
The angle through which the wheel turned can be calculated by applying the following kinematic equation;
[tex]\omega_f^2 = \omega_i^2 + 2\alpha \theta\\\\\theta = \frac{\omega_f^2\ -\ \omega_i^2}{2\alpha } \\\\\theta = \frac{(72)^2\ -\ (33.3)^2}{2(2.15)}\\\\\theta = 947.7 \ rad[/tex]
Therefore, the angle through which the wheel turned is 947.7 rad.
What effect does the velocity of a rotating object have on the centripetal force exerted on it?
Answer:
The velocity of a rotating object and the centripetal force exerted on it are proportional to each other. Centripetal force is equal to mv /r. If the velocity increases, the centripetal force increases as well.
Explanation:
The velocity has the direct effect on the centripetal force. Hence, increasing the velocity, centripetal force also increases or vice - versa.
The given problem is based on the basic concept of centripetal force and fundamentals involved in the centripetal force. The centripetal force is a center seeking force which comes to play, when any object undergoes the rotational motion around the circular path.
The mathematical expression for the centripetal force is given as,
[tex]F= \dfrac{mv^{2}}{r}[/tex]
Here,
m is the mass of an object.
v is the velocity of an object undergoing the motion around the circular path.
r is the radius of curvature.
From the concept of centripetal force, the velocity of a rotating object and the centripetal force exerted on it are proportional to each other. Centripetal force is equal to mv /r. If the velocity increases, the centripetal force increases as well.
Thus, we can conclude that the velocity has the direct effect on the centripetal force. Hence, increasing the velocity, centripetal force also increases or vice - versa.
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By calculating its wavelength (in nm), show that the second line in the Lyman series is UV radiation.
Answer:
λ = 102.78 nm
This radiation is in the UV range,
Explanation:
Bohr's atomic model for the hydrogen atom states that the energy is
E = - 13.606 / n²
where 13.606 eV is the ground state energy and n is an integer
an atom transition is the jump of an electron from an initial state to a final state of lesser emergy
ΔE = 13.606 (1 / [tex]n_{f}^{2}[/tex] - 1 / n_{i}^{2})
the so-called Lyman series occurs when the final state nf = 1, so the second line occurs when ni = 3, let's calculate the energy of the emitted photon
DE = 13.606 (1/1 - 1/3²)
DE = 12.094 eV
let's reduce the energy to the SI system
DE = 12.094 eV (1.6 10⁻¹⁹ J / 1 ev) = 10.35 10⁻¹⁹ J
let's find the wavelength is this energy, let's use Planck's equation to find the frequency
E = h f
f = E / h
f = 19.35 10⁻¹⁹ / 6.63 10⁻³⁴
f = 2.9186 10¹⁵ Hz
now we can look up the wavelength
c = λ f
λ = c / f
λ = 3 10⁸ / 2.9186 10¹⁵
λ = 1.0278 10⁻⁷ m
let's reduce to nm
λ = 102.78 nm
This radiation is in the UV range, which occurs for wavelengths less than 400 nm.
Suppose that the clay balls model the growth of a planetesimal at various stages during its accretion. Choose the planetesimal that is most likely
to pull in debris particle A.
Answer:2
Explanation:
Suppose that the clay balls model the growth of a planetesimal at various stages during its accretion, the planetesimal that is most likely to pull in debris particle A would be option B.
What is science?Science is the methodical, empirically-based pursuit and application of knowledge and understanding of the natural and social worlds.
Science is a way of learning about the world.
People may contribute to the development of new knowledge through science and utilize it to promote their objectives.
It is a method, a thing, and an organization all at once.
If the clay balls represent the formation of a planetesimal at different phases of its accretion, then option B represents the planetesimal that is most likely to draw in debris particle A.
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Find the lowest two frequencies that produce a maximum sound intensity at the positions of Moe and Curly.
Answer:
hello your question has some missing parts below is the complete question
and the missing diagram
The two speakers emit sound that is 180° out of phase and of a single frequency,ƒ, Find the lowest two frequencies that produce a maximum sound intensity at the positions of Moe and Curly.
answer : 1316.2 hertz
Explanation:
The frequency that produce the maximum sound intensity can be calculated using the relation below
dsin ∅ = n A
where A = dsin ∅ / n when n = 1 . d = 0.800
A = 0.800 * ( 1 / 3.162 )
A = 0.253 m
speed of sound = 333 m/s
frequency = speed / A
= 333 / 0.253 = 1316.2 hertz
how high off the ground is a book that has 120 J of potential energy and a 4-kg mass ?
Answer:
3 m
Explanation:
PE = mgh
120 J = (4 kg) (10 m/s²) h
h = 3 m
The book is 3m high from the ground.
What is potential energy?The energy by virtue of its position is called the potential energy.
PE = mgh
where, g = 9.81 m/s²
Given is the mass of book m =4kg and the potential energy P.E = 120J, then the height is
120 J = (4 kg) (10 m/s²) h
h = 3 m
Thus, the book is 3m high from the ground.
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During an earthquake, the top of a building oscillates with an amplitude of 30cm at 1.5Hz.
Part A What is the magnitude of the maximum displacement from the equilibrium of the top of the building? (in cm)
Part B What is the maximum velocity of the top of the building?(in cm/s)
Part C What is the maximum acceleration of the top of the building?(in cm/s^2
A.30cm
B.282.7cm/s
C.26655cm/s²
Explanation:
A. Maximum displacement= Amplitude= 0.3m
B.mum velocity=
2πAf= 2π x 1.5x30 =282.7cm/s
C. Maximum acceleration
= ( angular velocity)² A= (2πf)² A
=( 2 x 3.142 x 1.5)² x 30
2665.5cm/s²
LED light bulbs are more energy efficient, last longer, cost more, and have a smaller environmental impact from manufacture to recycling or disposal. The question is which costs less: incandescent, compact fluorescent (CFL), or LED lights?
(a) Ignore the time value of money. What is the average cost per hour for 25,000 hours of light-mg in a community with exist0.15/kWh electricity: for each choice?
(b) If the light is on 5%, 30%, or 60% of the time, how long until does it take for the 25,000 hours?
(c) Which type of light should a homeowner buy?
Answer:
a) Cost_incandescent = $ 375 , Cost _CFL = $ 75 , Cost_LED = $ 45
b) total = 20833 days , total_day = 3472 days , Total_days = 1736 d
c) LED
Explanation:
This problem is interesting to solve it, we must have a very important factor, such as the equivalence of the light emitted by the three systems, let's perform the calculation for a light intensity of 1000 lumen in the three bulbs.
consumed potence
incandescent 100W = fluorescent (CFL) 20W = Led 12 W
Now if we can solve your exercise
a) Let's find the power consumed by each type of bulb in the 25000 h
E = P t
incandescent
E_incandescent = 100 25000
E_incandescent = 2.5 105 Wh
E_incandescent = 2.5 103 kWh
Fluorescent (Low Power) CFL
E_CFL = 20 25000
E _DFL = 5 102 kWh
LED
E_LED = 12 25000
E_LED = 3 102 kwh
Let's use a direct rule of proportions (rule of three) for the cost of energy, if 1 kWh costs $ 0.15, the energy calculates how much it costs
incandescent
Cost_Incandescent = 2.5 103 kW / h ($ 0.15 / 1kWh)
Cost_incandescent = $ 3.75 102 = $ 375
CFL
Cost_CFL = 5 102 0.15
Cost _CFL = $ 75
LED
Cost_LED = 3 102 0.15
Cost_LED = $ 45
Here we do not take into account the possibility of burning incandescent bulbs and there is an extra replacement cost
b) for this part we use direct proportion rules
If the day has 24 hours and the light is on for 5%, how many days are 25000 hours?
for 5% = 0.05
hours on in a day
#_hoursdays = 24 hrs 0.05 = 1.2 hrs
total days
total = 25000 / 1.2
total = 20833 days
for 30% = 0.30
# _hoursday = 24 0.30 = 7.2 h
Total_days = 25000 h (1 day / 7.2 h)
total_day = 3472 days
for 60% = 0.602
#_hours = 24 0.60 = 14.4 h
Total _days = 25000h (1day / 14.4 h)
Total_days = 1736 d
c) you should decide between the CFL and the Led are much lower.
Between these two should take into account the cost of the bulbs, if we only take into account the cost of the energy consumed, the selection should be LED
A truck pulls a block 8 meters across a level surface at a force of 216 N over the course of 12 seconds. How much power did the truck use
Answer:
work = 1728
Power = 134
Explaination:
by using the formula,
Work(W)= Force(F)×Distance(D)
andPower(P)= Work(W)/Time taken(T)
In empty space, which quantity is always larger for X-ray radiation than for a radio wave?
a) amplitude
b) wavelength
c) frequency
d) speed.
Answer:
C
Explanation:
If you cook a marshmallow a few feet above a camp fire, it will cook mostly
by
A) radiation
OB) conduction
C) convection
Answer:
A) radiation.
Explanation:
Hello,
In this case, heat transfer from a body with a high temperature to a body with a lower temperature may occur when the bodies are not in direct physical contact with each other or when they are separated in space and it is is called heat radiation. Solids, liquids, or gaseous substances are able to emit energy via a process of electromagnetic radiation because of vibrational and rotational movement of their molecules and atoms.
Hence, since the camp fire is not in contact with the marshmallow and it still cooks it, we can infer that the mechanism of heat transfer is A) radiation. Moreover, both conduction and convection occur when the bodies are in contact.
Regards.
Parallel rays of monochromatic light with wavelength 582 nm illuminate two identical slits and produce an interference pattern on a screen that is 75.0 cm from the slits. The centers of the slits are 0.640 mm apart and the width of each slit is 0.434 mm. If the intensity at the center of the central maximum is 4.40×10^−4 W/m^2.
Required:
What is the intensity at a point on the screen that is 0.710 mm from the center of the central maximum?
Answer:
I = 0.56 10⁻⁴ W / m²
Explanation:
Double-slit interference is described by the expression
d sin θ = m λ
if we use trigonometry we can find the angle
tan θ= y / L
how the angles are small
tan θ = sin θ / cos θ = sin θ
sin θ = y / L
we substitute
d y / L = m λ
if we take into account that each slit also produces a diffraction phenomenon, the intensity distribution is the product of the intensity of the slits by the intensity of the diffraction process
I = I₀ cos² (d a) [(sin ba) / ba]²
a = π sin θ /λ
where the separation of the slits and b is the width of the slits
we substitute
I = I₀ [cos (d a)]² [sin ba /ba]²
a = π y / (L λ)
let's reduce the magnitudes to the SI system
λ = 582 nm = 582 10⁻⁹ m
L = 75.0 cm = 75.0 10⁻² m
d = 0.640 mm = 0.640 10⁻³ m
b = 0.434 mm = 0.434 10⁻³ m
Io = 4.40 10⁻⁴ W / m²
let's calculate for y = 0.710 mm = 0.710 10⁻³ m
a = π 0.710 10⁻³ / (75 10⁻² 582 10⁻⁹)
a = 5.11004 10³
I = 4.40 10⁻⁴ [cos (0.640 10⁻³ 5.11004 10³)]² (sin (0.434 10⁻³ 5.11004 10³) / (0.434 10⁻³ 5.11 10³)²
I = 4.40 10⁻⁴ cos² (3.2704) (sin 2.2178 / 2.2178)²
remember that the angles are in radians
I = 4.40 10⁻⁴ 0.9835 (0.79789 / 2.2178)²
I = 4.40 10⁻⁴ 0.9835 0.1294
I = 0.56 10⁻⁴ W / m²
Which step of the scientific method do you perform after you state the
problem?
O A. Collect data and observations.
B. Conduct an experiment.
C. Form a hypothesis.
D. Draw conclusions.
Answer:
form a hypothosis
Explanation:
A long, straight wire of radius R carries a steady current I that is uniformly distributed through the cross section of the wire. Calculate the magnetic field a distance r from the center of the wire in regions r ≥ R and r < R.
Answer:
a
When [tex]r \ge R[/tex]
[tex]B = \frac{ \mu_o * I}{ 2 \pi r }[/tex]
b
When [tex]r< R[/tex]
[tex]B = [\frac{\mu_o * I }{ 2 \pi R^2} ]* r[/tex]
Explanation:
From the question we are told that
The radius is R
The current is I
The distance from the center
Ampere's law is mathematically represented as
[tex]B[2 \pi r] = \mu_o * \frac{I r^2 }{R^2 }[/tex]
[tex]B = \frac{ \mu_o}{2 \pi } * \frac{r}{R^2}[/tex]
When [tex]r \ge R[/tex]
=> [tex]B = \frac{ \mu_o * I}{ 2 \pi r }[/tex]
But when [tex]r< R[/tex]
[tex]B = [\frac{\mu_o * I }{ 2 \pi R^2} ]* r[/tex]
A cubic container holds 2,350 g of fresh water. Use density and the volume formula below to
compute the dimensions of the tank.
V = 93
Density of Water: 1.00 g/cm3
O 13 cm^3
13.20 cm
O 13.3 cm^3
13.3 cm
O 13 cm
Answer:
Each side is 13.3 cm long
Explanation:
Use the formula for density as the mass per unit of volume, to find the volume of the tank:
[tex]density=\frac{mass}{Vol} \\Vol=\frac{2350}{1} \,\,cm^3\\Vol=2350\,\,cm^3[/tex]
Now we can find the dimensions of the tank knowing that it is cubic, and therefore its volume is [tex]side^3[/tex]:
[tex]side^3=2350\,\,cm^3\\side=\sqrt[3]{2350} \,\,cm\\side\aprox13.3\,\,cm[/tex]
Which of the following is not a way that astronomers can find how much dark matter there is in cluster of galaxies?
a. Using gravitational lensing effects to make a map of how much mass a cluster of galaxies contains.
b. Observe the radio waves coming from all dark matter; from the strength of the radio waves from each cluster, estimate the amount of dark matter needed to produce them.
c. Observing X-rays from the hot gas in the cluster, and using the amount of x-ray radiation to estimate the mass of the cluster.
d. Measuring how much slower galaxies in a rich cluster or supercluster are moving than the Hubble Law would predict.
e. Measuring the speeds with which the galaxies are moving around, to estimate the mass of the cluster needed to keep them from moving away.
Answer:
b. Observe the radio waves coming from all dark matter; from the strength of the radio waves from each cluster, estimate the amount of dark matter needed to produce them.
Explanation:
The universe is thought to be made up of 85% dark matters. Dark matter is called dark because it does not appear to interact with the electromagnetic field, which means it doesn't absorb, reflect or emit electromagnetic radiation, and is therefore difficult to detect. This means that option b is wrong since radio wave is an electromagnetic wave. Dark matter is a form of matter that makes up about a quarter of the total mass–energy density of the universe. Dark matter was theorized due a variety of astrophysical observations and gravitational effects that cannot be explained by accepted theories of gravity unless there were more matter in the universe than can be seen.
Water flows through a valve with inlet and outlet velocities of 3 m/s. If the loss coefficient of the valve is 2.0, and the specific weight of water is 9800 N/m3, the pressure drop across the valve is most nearly:
Answer:
The value is [tex]\delta P = 9000 \ Pa[/tex]
Explanation:
From the question we are told that
The inlet and outlet velocity is [tex]v = 3 \ m/s[/tex]
The loss coefficient is [tex]k = 2[/tex]
Th specific weight of water is [tex]w = 9800 \ N/m^3[/tex]
Generally the pressure drop across the valve is mathematically represented as
[tex]\delta P = \frac{1}{2} * k * \rho * v^2[/tex]
Here [tex]\rho[/tex] is the density which is mathematically represented as
[tex]\rho = \frac{w}{g}[/tex]
=> [tex]\rho = \frac{9800}{9.8}[/tex]
=> [tex]\rho = 1000 \ kg/m^3[/tex]
So
[tex]\delta P = \frac{1}{2} * 2* 1000 * 3^2[/tex]
[tex]\delta P = 9000 \ Pa[/tex]
Which of these changes will decrease the acceleration of an object? Select the two correct answers
increasing the applied net force
decreasing its mass
increasing its mass
decreasing the applied net force
Answer:
✔️From Newton's second law of motion we know that
✔️acceleration is inversely proportional to the mass of the body
✔️Here in this question increasing the mass of body will decrease its acceleration.
Explanation:
Hope it will help you :)
A 5.0 V battery storing 43.0 kJ of energy supplies 1.5 A of current to a circuit. How much energy does the battery have left after powering the circuit for 1.0 h? A. 43 kJ B. 16 kJ C. 41 kJ D. 27 kJ
Answer:
B. 16 kJ
Explanation:
Energy = VIt.............. Equation 1
Where V = Voltage, I = Current, t = time
Given: V = 5.0 V, I = 1.5 A, t = 1 h = 3600 s.
Substitute these values into equation 2
E = 5.0(1.5)(3600)
E = 27000 J
E = 27 kJ.
Amount of energy left = 43 kJ - 27 kJ
Amount of energy left = 16 kJ.
Hence the right option is B. 16 kJ
Complete the following statement: (In Boyle's law), the volume of a gas ________ when the ________ decreases.
Answer:
In Boyle's law the volume of a gas increases when the pressure decreases.
Explanation:
Boyle's law states that at fixed mass of gas, the volume of a gas is inversely proportional to the gas pressure, provided temperature remain constant.
Mathematically it be expressed as;
P₁V₁ = P₂V₂
As the volume of a gas increases, the pressure decreases, or as the volume of the gas decreases, the pressure of the gas increases.
Therefore, in Boyle's law the volume of a gas increases when the pressure decreases.
A cylindrical can of radius R is rolling across a horizontal surface without slipping.
A. After one complete revolution of the can, what is the distance that its center of mass has moved.
B. Would this distance be greater or smaller if slipping occurred?
Answer:
a) 2πR
b) greater
Explanation:
Radius of the cylinder = R
a) after one complete revolution, the center of mass would have moved a distance equal to the circumference of the radius formed by the cylinder.
distance moved = 2πR
b) If slipping occurred, there will be some distance lost during rotation, which will increase the distance that the center of mass will have to move.
Calculate the period of a satellite orbiting the Moon, 98 kmkm above the Moon's surface. Ignore effects of the Earth. The radius of the Moon is 1740 kmkm.
Answer:
3.6*10^18s
Explanation:
To find the period of the satellite
We need to apply kephler's third law
Which is
MP² = (4π²/G) d³
d=semi-major axis which is the distance from center of moon = 98km+1740km = 1838km
where M= mass of the moon = 7.3x10^22kg
P=period
G=newtonian gravatational constant= 6.67x10^-11
To find the Period solve for P
P = √[(4π²/G M)xd³]
P=√(4 π²/6.67x10^-22*7.3x10^22kg) x (1.838x10^6m)³]
= 3.6*10^18s
From what height must an oxygen molecule fall in a vacuum so that its kinetic energy at the bottom equals the average energy of an oxygen molecule at 300 K ?
Answer:
The value is [tex]h = 11930 \ m[/tex]
Explanation:
From the question we are told that
The temperature is [tex]T = 300 \ K[/tex]
Generally the root mean square speed of the oxygen molecules is mathematically represented as
[tex]v = \sqrt{\frac{3 * R * T }{M} } = \sqrt{ 2 * g * h }[/tex]
Here R is the gas constant with a value [tex]R = 8.314 \ J\cdot K^{-1} \cdot \ mol^{-1}[/tex]
M is the molar mass of oxygen molecule with value [tex]M = 0.032 \ kg /mol[/tex]
So
[tex]\frac{3 * 8.314 * 300 }{0.032} = 2 * 9.8 * h[/tex]
=> [tex]h = 11930 \ m[/tex]
What quantity, combined with universal physical quantities, can be used to determine a particle's de Broglie wavelength
Answer:
The quantity that will be combined to determine a particle's de Broglie wavelength is ENERGY.
Explanation:
Energy of a particle is given by;
E = hf
Where;
f is frequency of the particle's wave = c/λ
[tex]E = \frac{hc}{\lambda}[/tex]
[tex]\lambda= \frac{hc}{E}[/tex]
Where;
λ is the particle's de Broglie wavelength
h is the Planck's constant = 6.626 x 10⁻³⁴ J/s (This is one of the universal physical quantities)
c is the speed of light, = 3 x 10⁸ m/s (This is also one of the universal physical quantities)
E is the energy of the particle.
Therefore, the quantity that will be combined to determine a particle's de Broglie wavelength is ENERGY.
The de Broglie wavelength of a particle can be calculated using energy and universal physical parameters.
Define wavelength.The wavelength of a wave signal conveyed in space or down a wire is the distance among identical points (consecutive crests) in adjacent cycles.
When two persons have the same overall mentality and hence can communicate successfully, this is an instance of wavelength.
The de Broglie wavelength of a particle can be calculated using energy and universal physical parameters.
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Imagine that you are seeking treatment from a provider who offers complementary, alternative, or integrative treatment. Write at least one (1) question you would want to ask the provider about their education or credentials.
Explanation:
It is important to reach a decision only after noting the quality of responses received about the following pertinent questions;
1. What training program(s) have they gone through?
2. Do they have any licences?
3. Have they received any certifications from a recognized body?
4. Do they have any ongoing or past scientific research work?
5. Awards they've received, if any?
As this is a key decision of someone's life, making the decision on the basis of the provider's experience with similar cases and quality of service is the most important.
We should ask the following questions to the provider,
1. Does the practitioner have licenses for such practice?
2. Where did the practitioner receive their educational qualifications from and do they still learn to be up to date with the advanced treatments available in the market?
3. Was the practitioner awarded for any similar cases or handled similar cases?
4. For how long they have been practicing?
5. Does the practitioner have a membership of any organizations or professional societies?
6. Can the treatment be covered by the insurance?
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